Foundations — Angle of friction, angle of repose — derivation

Before you can derive that result in the parent topic, you need every letter and picture the derivation quietly assumes. This page builds them one at a time, from nothing. If a symbol appears in the parent, it is defined here first.

1. What "force" means, and the arrow that draws it

Look at figure 1. The block just sits on the floor, yet three arrows act on it. We must learn to read every arrow before we can add them.

The unit of force is the newton, written $\text{N}$. One newton is roughly the weight of a small apple resting in your hand.

2. Weight $mg$ — gravity's downward arrow

• $m$ = mass (kg)
• $g$ = gravitational field strength $\approx 10\text{N/kg}$ (we use $10$ for easy arithmetic; the real value is $9.8$)
• $mg$ = weight, an arrow pointing straight down, always.

Why the topic needs this: on a tilted ramp, gravity is the only thing trying to make the block slide. Every "driving force" in the derivation comes from $mg$.

3. Normal force $N$ — the surface pushing back, perpendicular

In figure 1 the cyan arrow labelled $N$ points straight up out of the floor. On a tilted ramp it points straight out of the ramp — still perpendicular to the surface, so it leans along with the ramp (figure 4).

Why the topic needs this: $N$ measures how hard the block and surface are squeezed together — and friction's strength is built directly on $N$.

4. Friction force $f$ — the sideways grip

Read that carefully: friction is not a fixed number. It is lazy — it supplies only as much grip as needed, right up to a maximum of ${\mu }_{s}N$. Push harder than that and the block breaks free.

Why the topic needs this: the whole derivation lives at the moment $f={\mu }_{s}N$. That single equation is the engine of both results.

5. The coefficient of friction ${\mu }_s$ — the "grip number"

${\mu }_s$ is defined by the limit equation itself: ${\mu }_s=\frac{f_{\max }}{N}$ It is the ratio "maximum sideways grip ÷ how hard you're pressed together". Because it is a ratio of two forces, the newtons cancel and ${\mu }_s$ is just a bare number.

Why the topic needs this: the punchline of the whole topic is $\text{angle}=\arctan ({\mu }_s)$. Everything funnels into this one number.

6. The right triangle, and what $\tan$ measures

The derivation turns forces into an angle. The bridge between "two arrows" and "one angle" is the right triangle — so we must define it and the tool that reads its steepness.

Notice a key fact from figure 2: as $\theta$ grows from $0°$ toward $90°$, the opposite side stretches while the adjacent stays put, so $\tan \theta$ climbs from $0$ upward without bound. Every steepness has its own tangent — that one-to-one link is what lets us go backward.

7. $\arctan$ — undoing the tangent

Think of it as the question "which angle has this tangent?"

Why the topic needs this: the derivation lands on $\tan (\text{angle})={\mu }_s$. To get the angle out, we must undo the tangent — that is precisely $\arctan$. This is the final key.

8. Resolving a vector into components

Gravity points straight down, but on a tilted ramp the ramp's "along" and "perpendicular" directions are tilted too. To use them, we split the single weight arrow into two arrows on those tilted axes.

On a ramp tilted at angle $\theta$, the downward weight $mg$ splits into:

• Along the slope (trying to slide the block down): $mg\sin \theta$
• Perpendicular to the slope (pressing into the ramp): $mg\cos \theta$

Why the topic needs this: the repose derivation compares $mg\sin \theta$ (the slide-driver) against friction built from $mg\cos \theta$ (the press). Without resolving, we can't write those two forces.

9. Equilibrium — the balanced-arrows condition

For a block sitting still on a ramp, equilibrium along each axis separately gives us:

• Perpendicular: $N=mg\cos \theta$ (surface push balances the pressing component)
• Along slope, at the verge: $mg\sin \theta =f_{\max }={\mu }_{s}N$

Why the topic needs this: equilibrium is the licence to write "these forces are equal". Every equation in the derivation is really an equilibrium statement.

10. Putting it together — the two named angles

Now every symbol exists, so we can state (not yet fully derive) what the parent proves:

Figure 4 shows the total reaction $R$ — the single arrow you get by adding $N$ and $f$ — leaning by angle $\lambda$ from the normal. That is the picture the parent's Pythagoras step builds.

The same idea, $\arctan (\mu )$, returns in Banking of Roads for the safe-speed tilt of a curved road — so mastering it here pays off twice.

Prerequisite map

Equipment checklist

Test yourself — cover the right side and answer before revealing.